Dynamical Systems 2, MA 761

Dynamical Systems 2, MA 761 Topological Dynamics This material is based upon work supported by the National Science Foundation under Grant No. 9970363 1 Periodic Points 1 The main objects studied in the theory of dynamical systems are maps f : X X, where X is a space with some structure. For instance, this structure can be topological, differentiable, or there may be a measure on X. We work with the first case mostly. That is, let f : X X be a continuous map of a compact metric space X into itself. We consider the iterates of f, defined by induction: f 0 = id X, and then f n+1 = f n f. Thus, f n = f f. This means that we look at the set of all iterates of f, {f n } }{{} n Z+. n times The special case which is considered as often as the general one (and maybe even more often) is when the map f is invertible. We require then that f 1 is also within the class of maps we consider. That means that f and f 1 are continuous; then f is called a homeomorphism. If we consider an invertible f then we study also negative iterates of f, defined as f n = (f 1 ) n = (f n ) 1. EXAMPLE 1.1. Let X = [0, 1] and let f : X X be given by the formula { 2x if 0 x 1/2, f(x) = 2 2x if 1/2 x 1. The map f is called the (full) tent map. Call x a fixed point iff f(x) = x. Call x a periodic point iff f p (x) = x for some p > 0; all such p are called periods of x. If p is the minimal positive number such that f p (x) = x say that p is the minimal period of x. CLAIM 1.2. The periods of x are {mp : m 0} where p is the minimal period of x. Proof. Otherwise there exists n = mp + i, 0 < i < p such that f n (x) = x. However, f mp (x) = x too, so f i (x) = f i (f mp (x)) = x, a contradiction with the choice of p. Observe that in the arguments we do not use the continuity of f or any other specifics of the situation. Therefore the proof remains valid for a map T : Y Y of a set Y into itself, that is on the set-theoretic level. 1

EXAMPLE 1.3. Identity map. A map f : X X is called periodic if f p = id X for some p > 0; then p is called a period of f which allows one to define the minimal period of f. By Claim 1.2 and the observation after that all periods of a map f are multiples of its minimal period which remains true also on the set-theoretic level. EXERCISE 1.4. Describe all interval homeomorphisms. Prove that all periodic maps are homeomorphisms and then describe periodic interval maps. In particular, show that their possible minimal periods are only 1 and 2. Moreover, show that if all points of the interval are periodic then the map is periodic too. EXAMPLE 1.5. A map f : [0, 1] [0, 1] with f(x) = bx(1 x), 0 b 4. EXAMPLE 1.6. Rotations of the circle, rational and irrational. Notice that irrational rotations of the circle have no periodic points while rational rotations by an angle 0 p/q < 1, p, q coprime, are periodic of period q. EXAMPLE 1.7. Shifts of the plane, reflections with respect to a straight line. A reflection of the plane is periodic of period 2 while a shift by a non-degenerate vector has no periodic points at all. CLAIM 1.8. The set Per 1 (f) = Fix(f) of all fixed points of f is closed. Proof. If not then there exists a sequence of fixed points x n x such that x is not fixed. However this contradicts the continuity of f. COROLLARY 1.9. The set Per p (f) of all points of period p is closed; the set of all points of period at most p is closed too. Also, Per p (f) can be described as the set of all points whose minimal period is a divisor of p. Proof. The second half follows from 1.2. To prove the first half, apply 1.8 to f p. EXAMPLE 1.10. The following map is an example of a map for which the set of all points whose minimal period is 2, is not closed: f : [0, 1] [0, 1], f(x) = 1 x. A point x is called (pre)periodic if it is mapped by some power of f into a periodic point. The sequence {f i (x), i 0} is called the trajectory of x while the set n=0f n (x) = orb(x) is called the orbit of x. EXERCISE 1.11. A point x is (pre)periodic iff orb(x) is finite. 2

2 Periodic Points 2 We begin by solving some of the exercises. First we need the following simple claim. CLAIM 2.1. Let h : [0, 1] [0, 1] be a homeomorphism. Then one of the following holds: (1) h(0) = 0, h(1) = 1, any interval (a, b) complementary to the set Fix(h) is mapped onto itself by f in such a way that all the points inside (a, b) are mapped to the right of themselves, or all the points inside (a, b) are mapped to the left of themselves, and so Per(h) = Fix(h); (2) h(0) = 1, h(1) = 0, there exists a unique fixed point a (0, 1), and the set of all intervals complementary to the set Per 2 (h) can be partitioned into pairs of intervals (b, c) and (d, e) such that h[b, c] = [d, e], h[d, e] = [b, c], h(b) = e, h(e) = b, h(c) = d, h(d) = b and there are no periodic points inside (b, c) (d, e), and so the set Per(h) of all periodic points of h equals the set Per 2 (h) of all points of period 2. Proof. Since h is a homeomorphism, either h(0) = 1, h(1) = 1 or h(0) = 1, h(1) = 0. Consider the cases. (1) h(0) = 1, h(1) = 1. Consider the set Fix(h). This is a closed subset of [0, 1]. If it does not coincide with [0, 1] then on any interval (a, b) complementary to Fix(f) in [0, 1] points are mapped either to the left or to the right because otherwise there will be fixed points inside (a, b), a contradiction. Moreover, because h is 1-to-1, h(a, b) = (a, b). Therefore, there are no periodic points inside (a, b) - all points inside (a, b) are mapped more and more to the right or to the left all the time. (2) h(0) = 1, h(1) = 0. There must be a fixed point a (0, 1). Then h[0, a] = [a, 1] and h[a, 1] = [0, a]; in other words, the map flips the interval over the point a. Then for g = h 2 we have g(0) = 0, g(1) = 1, g(a) = a, so the previous case applies (obviously, iterates of a homeomorphism are homeomorphisms). If we go back to h we will see, that Per(f) = {a} B where B is the set of all points of minimal period 2. An interval J complementary to Per(h) is such that J and h(j) are disjoint while h 2 (J) = J. Also, all points inside J are mapped into the same direction by h 2. The same holds for every interval complementary to Per(f h). This concludes the description of homeomorphisms of [0, 1]. Now we can do Exercise 1.4 whose part the above claim actually is. 3

EXERCISE 1.4. Describe all interval homeomorphisms. Prove that all periodic maps are homeomorphisms and then describe periodic interval maps. In particular, show that their possible minimal periods are only 1 and 2. Moreover, show that if all points of the interval are periodic then the map is periodic too. Proof. Observe that in general if all points of a space are periodic for a map T : X X, then this does not necessarily imply that T is a periodic map because minimal periods of T -periodic points may well be unbounded. It is not difficult to give an example of such a map on a specifically designed compact space. Indeed, consider a compact space X which is a sequence of points x i on the interval [0, 1] converging to 1 together with the point 1 itself. Divide {x i } into finite subsets P 0, P 1,... so that their cardinalities increase to infinity and P 0 < P 1 <... in the natural sense. Define a map T : X X so that P i is invariant and, moreover, any P i is a periodic orbit of T. Then it is easy to see that T : X X is not periodic while all points of X are periodic. So, the last question of the exercise makes sense. Observe however that if f : X X is such that all points are periodic then f must be a homeomorphism. Indeed, denote the minimal period of a point x X by p x. Then f is surjective because x = f(f px 1 (x)) for any x. Also, if f(x) = f(y) = z then z belongs to the periodic orbits of x and y which implies that these two orbits are the same. Denote their common period by p. Then f p (x) = f p (y) = f p 1 (z) = x = y, so f is injective. So, a map f whose all points are periodic is a 1-to-1 continuous map of a compact metric space onto itself. Therefore f is a homeomorphism (that is, its inverse is continuous too). Let us go back to the original question of the description of the interval maps whose all points are periodic. As follows from the previous paragraph, such maps are homeomorphisms. Then Claim 2.1 implies that the only two ways a homeomorphism can be a map whose all points are periodic are as follows: (1) in the case (1) of Claim 2.1 the map f must be identity; (2) in the case (2) of Claim 2.1 the map f must be such that f 2 is the identity and in fact there exists the unique fixed point a [0, 1] such that f[0, a] = [a, 1], f[a, 1] = [0, a] and all other points of the interval are of period 2. Let us now do Exercise 1.11. EXERCISE 1.11. A point x is (pre)periodic iff orb(x) is finite. Proof. It is enough to check that if orb(x) is finite then x is (pre)periodic. Indeed, by the assumption there exist two numbers i > 0, j > 0 such that f i (x) = f i+j (x), hence f i (x) is periodic and we are done. 4

3 One example related to symbolic dynamics EXAMPLE 3.1. Let X = [0, 1] and let f : X X be given by the formula 3x if x L = [0, 1/3], f(x) = 2 3x if x N = [1/3, 2/3], 3x 2 if x R = [2/3, 1]. This is so-called saw bimodal map. Our space is a metric compact, and the map is continuous. Now, consider the set A of all points x whose orbits are contained in J = L R. Show that A is the standard 1/3-Cantor set and study f A. The idea is to construct the set A step by step. First consider the set A 0 of all points x which belong to J. Then consider all points x such that x and f(x) J, etc. Let us pass on to a detailed proof. Proof. We begin by a very important construction. Let S be a finite set consisting of more than one point, for instance S = {1, 2,..., s} with s > 1. Define Σ = S and Σ + = 0 S (for brevity we skip the dependence upon s in our notation here). More precisely, Σ = i= S i and Σ + = i=0 S i, where S i = S for each i. Thus, the elements of Σ are the doubly infinite sequences (..., x 2, x 1, x 0, x 1, x 2,...) with x i S for all i, and the elements of Σ + are the usual one-sided sequences (x 0, x 1, x 2,...) with x i S for all i. We will mostly work with Σ +. We define a shift σ on Σ and Σ + (we will use the same letter in both cases) as the shift by one to the left. This means that σ(x 0, x 1, x 2,...) = (x 1, x 2, x 3,...). To write the formula for σ on Σ is more difficult. For this we have to introduce notation for the points of Σ which shows where the 0-th coordinate is. Namely, we shall write x = (..., x 2, x 1, * x 0, x 1, x 2,...) if the 0-th coordinate of x is x 0. With this notation we can write σ(..., x 2, x 1, * x 0, x 1, x 2,...) = (..., x 1, x 0, * x 1, x 2, x 3,...). Moreover, in the case of the space Σ, σ is 1-to-1. In the case of the space Σ +, σ is s-to-one. We will regard Σ and Σ + as metric spaces. We need the following definition: a cylinder is a set of the form C y n,y n+1,...,y n 1,y n = {(..., x 2, x 1, x 0, x 1, x 2,...) Σ : x i = y i for all i { n, n + 1,..., n 1, n}} in the space Σ and of the form C y0,y 1,...,y n = {(x 0, x 1, x 2,...) Σ : x i = y i for all i {0, 1,..., n}} in the space Σ +. We define a metric in these spaces by setting d(x, y) = 2 k, where k is the smallest non-negative integer such that there is m with m = k and such that the m-th terms of the sequences x and y are different. Denote by S a two-symbol set S = {L, R} and also by Σ + the set of all one-sided sequences of elements of S. Thus, the elements of Σ + are the usual one-sided sequences 5

(x 0, x 1, x 2,...) with x i S for all i (for brevity we skip the dependence on s = 2). The map f maps each of the intervals L, R onto the whole [0, 1] in an increasing way. Now we apply the coding procedure to the system (A, f) with the partition {L, R} of J = L R. Denote the set of points x [0, 1] such that the point f i (x), 0 i n 1 lies in J, by X n 1. For any x X n 1 we look at the set L or R in which the point f i (x) lies for i = 0, 1, 2,..., n and call this set A i. In such a way we get for every point x A a code (A 0, A 1, A 2,...) Σ +, sometimes also called the itinerary of x. Observe, that while the finite segment of the sequence (A 0, A 1, A 2,..., A n 1 ) is defined for all points of X n 1, the infinite sequence (A 0, A 1, A 2,...) Σ + is defined for all x A only. We claim that given a code there is a unique point with this code. Let K = (A 0, A 1, A 2,...) Σ +. For a given n the set of points whose code begins with (A 0, A 1,..., A n 1 ) is equal to I n (K) = n 1 i=0 f i (A i ) (this immediately follows from the definitions). We show by induction that for every code K such a set is an interval of length 3 n whose homeomorphic f n -image is [0, 1]. This is definitely true for n = 1 because then the set of points whose code begins with L(R) is simply the interval L(R) itself. If it is true for some n then, since I n+1 (K) = A 0 f 1 (I n (σ(k))) and by the induction hypothesis I n (σ(k)) is an interval of length 3 n, we get that I n+1 (K) is an interval of length 3 n 1 whose homeomorphic f n+1 -image is [0, 1]. This completes the induction step. Now we have a descending sequence (I n (K)) n=0 of closed non-empty subsets of [0, 1]. Since [0, 1] is compact, the intersection of all I n (K) is non-empty. Since the length of I n (K) goes to 0 as n, this intersection consists of one point. We call this point ϕ(k). In such a way we define a map ϕ : Σ + A. Since every point of A has a code, this map is onto. On the other hand, since the itinerary of any point x A is welldefined, we conclude that this is a 1-to-1 map. We shall show that ϕ is continuous. Let lim n K n = M = (B 0, B 1, B 2,...). As n, longer and longer initial pieces of K n are the same as the initial pieces of M, so if we choose any m then I m (K n ) = I m (M) if n is sufficiently large. This implies that ϕ(k n ) ϕ(m) 3 m if n is sufficiently large. Therefore lim n ϕ(k n ) = ϕ(m), so ϕ is continuous. Look at I n (K). We know that f n (I n (K)) = [0, 1] and f n on I n (K) is linear and increasing. On the other hand, I n+1 (K) = f n (A n ) I n (K). Thus, depending on whether A n = L or A n = R, the interval I n+1 (K) is either the left of the right third of I n (K). This implies that A is constructed the same way as the standard Cantor set is, and thus A coincides with the standard Cantor set. If x, y A, d(x, y) ɛ then d(f n (x), f n (y)) 3 n (ɛ). So if d(x, y) < 3 n 1 then the first n entries of the itineraries of x and y coincide because f i (x), f i (y), 0 i n 1 cannot belong to L, R or R, L respectively. Thus, ϕ 1 is continuous. Hence ϕ is a homeomorphism and Σ + is compact. 6

4 Structural stability In math objects which look similarly are treated as the same object; the same is done in dynamical systems. Let X and Y be metric spaces and let f : X X and g : Y Y be continuous maps. If there is a homeomorphism h : X X with h f = g h (in other words, the diagram f X X h Y g commutes), we will say that f and g are conjugate. Notice that in this case X and Y are homeomorphic. The homeomorphism h is called a conjugacy (between f and g). In this situation, if h is not necessarily a homeomorphism, but just a continuous map of X onto Y, we say that g is a factor of f and h is a semiconjugacy of f with g. Another way of looking at the maps above is as follows: one of them can serve as a model for the other one providing a system of coordinate in which the original map is easier to deal with. Major questions in dynamical systems theory are whether two maps are conjugate or not and what extra properties the conjugacy ϕ has (smoothness and the like). Clearly, finding invariants defined in topological terms (topological invariants) is very important for figuring out if two maps are conjugate: if the value of an invariant is different for the two maps we consider they cannot by conjugate. The ideal situation is when one can come up with a complete collection of invariants so that two maps are conjugate if and only if they have the same invariants (which is why the system is called complete). A major topic in dynamical systems is to figure out how the dynamics of a map changes under small perturbations. Here small means small in the sense of a certain topology/distance introduced in the space of all maps. Usually and because C 0 -metric allows rather wild perturbations of a map, the metric which is considered is smooth and could be C n with n 1 (recall that two maps are C n -close if the maps and all their derivatives up to the n-th derivative are close). Now the above problem can be restated as follows: what properties does a map have to have in order to guarantee that its small perturbations are conjugate to itself? A map f which has a C n -neighborhood U such that all maps from U are conjugate to f is called C n -structurally stable. Similar definitions can be given if we restrict our attention onto homeomorphisms only. h Y EXAMPLE 4.1. Consider an orientation preserving C 1 -homeomorphism f : [0, 1] 7

[0, 1] such that f (0) > 1 > f (1) and there are no other fixed points of f and show that f is structurally stable in the class of homeomorphisms. Proof. Consider a map g which is very close to f in the C 1 -sense and show that f and g are conjugate. To do this let us show that if g is very close then g(0) = 0, g(1) = 1 and g (0) > 1 > g (1) and there are no other fixed points of g either. Indeed, the first part of the claim follows from the fact that g is a homeomorphism while the second follows from the choice of g very close to f in C 1 -topology. Let us show that 0, 1 are the only fixed points of g. To this end first choose small ε > 0 and δ > 0 such that for every x [0, ε] we have f (x) > 1 + δ and for every x [1 ε, 1] we have f (x) < 1 δ. Then choose g so close to f that g (x) > 1 + δ/2 for x [0, ε] and g (x) 1 δ/2 for x [1 ε/2, 1]. This implies that g has no fixed points inside J = [0, ε] [1 ε, 1] because otherwise there would be points z J with g (z) = 1, a contradiction. Now, since f [ε, 1 ε] has no fixed points then there exists γ > 0 such that f(x) x > γ for all ε x 1 ε]. If g is chosen close enough to f this implies that g(x) x > γ/2 for all ε x 1 ε] and completes the proof of the fact that g has only two fixed points, 0 and 1. It remains to prove that any two maps F and G of [0, 1] into itself such that the endpoints are fixed and the points inside map to the right are conjugate. To show that this is indeed the case let us choose a point x (0, 1) and denote [x, F (x)) by I. Then F (I) is an interval to the right of I whose left endpoint is F (x) (so it is attached to I from the right) and F 1 (I) is an interval to the left of I whose right endpoint is x (so it is attached to I from the left). The union of the sets F n (I), < n < covers the entire (0, 1) (to prove that one needs the assumption about the fixed points of F being only 0 or 1). Choose a point y (0, 1), consider the interval J = [y, G(y)) and perform the same construction as above but now for G and J. Construct an increasing homeomorphism ϕ : I J. Then extend it onto [0, 1] as follows: 1) for every point x (0, 1) choose the unique n x = n such that F n (x) I; 2) define ϕ(x) as G n ϕf n (x) (it is well-defined since F n (x) I). Thus, iterates F n (I) will be mapped onto their counterparts G n (J) in a way compatible with the way ϕ maps I onto J. Moreover, ϕ is continuous and can be extended onto [0, 1] by declaring that ϕ(0) = 0, ϕ(1) = 1. To finish the proof one needs to do Exercise 4.1 below. EXERCISE 4.1. Finish the proof, i.e., show that ϕ f = g ϕ. EXERCISE 4.2. For the class of increasing interval homeomorphisms of class C 1 whose all fixed points are such that the derivative at them has the absolute value greater than 1 or less than 1 describe the complete topological invariant. 8

5 Non-wandering Points Let X be a compact metric space and f : X X a continuous map. We call a point x X wandering if there exists a neighborhood U of x such that U f n (U) = for every n > 0. A point is called non-wandering if it is not wandering. The set of non-wandering points of f is denoted Ω(f). Recall that a point is called isolated if it is open as a set. The following proposition lists the basic properties of the set of non-wandering points. PROPOSITION 5.1. Let f : X X be a continuous map of a compact metric space X into itself. Then the following properties hold. (a) Ω(f) is compact and contains the closure of all periodic points of f. (b) Ω(f) is invariant, that is f(ω(f)) Ω(f). (c) If f is a homeomorphism then f(ω(f)) = Ω(f). (d) Ω(f n ) Ω(f) for any n > 0. (e) If f is a homeomorphism then Ω(f 1 ) = Ω(f). Proof. (a) By the definition, the set of wandering points is open. Therefore the set of non-wandering points is closed, and hence compact (since X is compact). (b) Assume that x Ω(f). Let U be a neighborhood of f(x). Then V = f 1 (U) is a neighborhood of x. Since x is non-wandering, there exists n > 0 such that V f n (V ). This means that there is y V such that f n (y) V. Then f(y) f(v ) = U and f n (f(y)) = f(f n (y)) f(v ) = U. Therefore U f n (U). Therefore, f(x) Ω(f). (d) Assume that x Ω(f n ). If U is a neighborhood of x then there exists m > 0 such that U (f m ) n (U). Since (f m ) n = f mn, this shows that x Ω(f). (e) Assume that f is a homeomorphism and x Ω(f). If U is a neighborhood of x then there exists n > 0 such that U f n (U). This means that there is y U such that f n (y) U. Therefore U f n (U). This shows that Ω(f) Ω(f 1 ). This is true for all homeomorphisms of X, in particular for f 1. Therefore Ω(f 1 ) Ω((f 1 ) 1 ) = Ω(f). (c) From (e) and (b) it follows that if f is a homeomorphism then f 1 (Ω(f)) = f 1 (Ω(f 1 )) Ω(f 1 ) = Ω(f), so Ω(f) f(ω(f)). Together with (b) this gives (c). 9

EXAMPLE 5.2. Consider the space X = {0, 2} { 2 n : n 0} {2 n : n 0} {2 + 2 n : n 0}. Define the map f : X X by 2x if x < 0, f(x) = x if 0 x 1, x 2 if x 2. Clearly, X is compact and f is continuous. All the points of X, except 0 and 2, are isolated and not periodic. Therefore they are wandering. The point 0 is a fixed point, so it is non-wandering. In any neighborhood of the point 2 there are points of the form 2 + 2 n. The image of such a point is 2 n and the 2n + 2-nd image of this point is 2. Therefore 2 is non-wandering. Thus, Ω(f) = {0, 2}. However, we have f({0, 2}) = {0}. This shows that the assumption in (c) that f is a homeomorphism is essential. The set U = {2} {2+2 n : n 0} is a neighborhood of 2. Set V = {0} { 2 n : n 0}. We have f 2 (U) V and f 2 (V ) V. Since V is disjoint from U, we get U f 2m (U) = for every m > 0. So, 2 is wandering for f 2, and in (d) we might not have the equality. Also, the set Ω(f Ω(f) ) (which is {0} here) can be smaller than Ω(f). The next proposition gives a reason why the notion of the set of non-wandering points is important. We will use the notation dist(x, y) for the distance between two points x and y, and dist(x, A) for the distance of a point x from the set A (that is, dist(x, A) = inf y A dist(x, y)). Proposition 5.3. Let f : X X be a continuous map of a compact metric space X into itself. Then every point is attracted by the set of non-wandering points, that is for every x X. lim dist(f n (x), Ω(f)) = 0 n Proof. Suppose that there is a point x X such that dist(f n (x), Ω(f)) does not tend to zero. Then there is an ε > 0 and a subsequence of the orbit (f n (x)) n=0 of x such that the distance of every point of this subsequence from Ω(f) is greater than or equal to ε. Since X is compact, there is a subsequence of this subsequence that converges to some point y X. Clearly, dist(y, Ω(f)) ε. For every neighborhood U of y there are arbitrarily large integers n with f n (x) U. We choose two of them, n < m. We have f m (x) U, and since f n (x) U, we get f m (x) = f m n (f n (x)) f m n (U). Thus, U f m n (U). This proves that the point y is non-wandering, contrary to the property dist(y, Ω(f)) ε > 0. This contradiction completes the proof. We can interpret the above proposition as follows. The trajectory of every point approaches Ω(f), so on longer and longer pieces it looks like trajectories of some points of Ω(f). Thus, all interesting dynamics can be found in Ω(f). 10

6 Limit Sets For any point x X we define the ω-limit (omega limit) set ω(x) of x (or of the trajectory of x) as the set of limits of all convergent subsequences of (f n (x)) n=0. Thus another way of stating Proposition 5.3 is that ω(x) Ω(f) for every x X. Let us list some properties of limit sets. EXAMPLE 6.1. Let x be a periodic point. Then ω(x) = orb(x); in particular, if a map is periodic then the limit set of any point is its periodic orbit. Similarly, if x is preperiodic and y orb(x) is periodic then ω(x) = orb(y). Let us prove the following simple but useful claim concerning limit sets. PROPOSITION 6.2. If ω(x) is a singleton then it is a fixed point. Proof. If f n (x) y then by continuity f(y) = y as desired. The cases described in Example 6.1 are not the only ones which describe the trajectories of points whose limit sets are periodic orbits. Another series of examples is presented in Example 4.1 and actually in earlier given Exercise 1.4. Indeed, we can easily prove the following proposition. PROPOSITION 6.3. For any orientation preserving interval homeomorphism f and any point x the limit set ω(x) is a fixed point; in the orientation reversing case ω(x) could be a fixed point or a periodic orbit of period 2. Proof. Without loss of generality we may assume that f : [0, 1] [0, 1] is orientation preserving so that f(0) = 0 and f(1) = 1. It is enough to consider a point x which is not fixed. Then x I where I = (a, b) is invariant under f, f(a) = a, f(b) = b and all points of I are mapped in the same direction. For the sake of definiteness let all y I be mapped to the right. Then x < f(x) <... and therefore ω(x) is the unique limit point of the increasing sequence (x, f(x),...). By Proposition 6.2 ω(x) is a fixed point which must belong to Ī and be greater than x. Hence ω(x) = b. PROPOSITION 6.4. Let f : X X be a continuous map of a compact metric space X into itself, x X. Then ω(x) is closed, f(ω(x)) = ω(x) and ω(x) Ω(f). Moreover, if the orbit of x is finite then x is preperiodic and ω(x) is a periodic orbit. Proof. The first claim follows from the fact that the set A of all limit points of any sequence {y i } in X is closed. Indeed, if z / A then there exists an open U z and N such that U {y i } i>n =. This implies that U A = and thus X \ A is open. Since 11

X is compact, ω(x). Observe that the notion of the limit set can be introduced in the non-compact case as well and in a non-compact metric space ω(x) may be empty; however it must be closed in any case (e.g., for the shift on the plane the limit sets of all points are empty). It is easy to see that ω(x) Ω(f). Indeed, if y ω(x) then for some {n k } we have f n k (x) y. If U is a neighborhood of y then there exists k such that f n k(x) U and f n k+1 U, hence f n k+1 n k(u) U and so y Ω(f) as desired. To prove that f(ω(x)) = ω(x) we prove that f(ω(x)) ω(x) and that f(ω(x)) ω(x). Indeed, if y ω(x) then y = lim k f n k (x) for some subsequence nk. Then by continuity of f we get f(y) = lim f n k+1 (x) ω(x) which implies f(ω(x)) ω(x). Observe that this holds in non-compact metric spaces. Now, consider a sequence f n k 1 (x). Since X is compact it has a limit point z - in other words, there exists a subsequence {m j } of {n k 1} such that f m j (x) z. Then f m j+1 f(z) and on the other hand since {m j +1} is a subsequence of {n k } we see that f(z) = y. This proves that f(ω(x)) ω(x). Now, if orb(x) is finite then we can find the smallest n for which there exists i such that f n+i (x) = f n (x) and then pick the smallest such i. Then it is easy to see that f n is a periodic point of the minimal period i and ω(x) = {f n (x),..., f n+i 1 (x)}. Denote the set of all periodic points of f by Per(f) and the union of all limit set of points of f by ω(f). Then Proposition 6.4 implies that Per(f)] Ω(f) and that Per(f) ω(f) Ω(f). Let us classify topological types of limit set for interval maps. PROPOSITION 6.5. Let f : [0, 1] [0, 1] be a continuous map. Then for any x either ω(x) is a nowhere-dense set or ω(x) is a finite collection of intervals cyclically permuted by f on which a map is surjective. Proof. Suppose that ω(x) is not nowhere dense. Then it has a component I which is a non-degenerate interval. Since I ω(x) we can choose two integers m, m + n such that f m (x) I and f m+n (x) I. Hence f n (I) I. On the other hand, I is a component of an invariant closed set ω(x). Therefore in fact f n (I) I. The union of intervals A = n 1 i=0 f i (I) is then a closed invariant set containing forward iterates of x which implies that ω(x) A. Since we know that A ω(x) we see that A = ω(x). To see now that A is a finite collection of intervals cyclically permuted by f choose the smallest m > 0 such that f m (I) I. Then since I is a component of A = ω(x) we have that f m (I) I, and by the minimality of m all smaller iterates of I are pairwise disjoint. Finally, since f A must be onto, f m (I) = I as desired. 12

7 Recurrent Points An important definition related to that of the limit set is that of a recurrent point. A point x is said to be recurrent if x ω(x). Let us list simple properties of recurrent points. PROPOSITION 7.1. A periodic point is recurrent whereas a preperiodic non-periodic point is not recurrent. PROPOSITION 7.2. Any recurrent point x belongs to Ω(f) (cf Proposition 6.4). PROPOSITION 7.3. If f is an interval homeomorphism then the only recurrent points of f are its periodic points. The notion of a recurrent point is an obvious generalization of that of a periodic point: in the latter case the point comes back exactly in itself while in the former case the point comes back close to itself with approximation becoming better and better. In some topologically defined cases recurrent points in fact have to be periodic. PROPOSITION 7.4. Isolated recurrent points are periodic. Proof. If x is isolated and recurrent then for small enough ε the fact that d(f n (x), x) < ε implies that f n (x) = x as desired. It is sometimes useful to consider a set {f j (x)} = orb x which is closely related to the limit set of x. Their relationship is described in the following lemma. LEMMA 7.5. orb x ω(x). Moreover, orb x = ω(x) if and only if x is recurrent. Proof. If y ω(x) then by the definition in any neighborhood U of y there is a point of orb x in U. Hence y orb x. Now, suppose that x is recurrent, that is x ω(x). Then all points of orb x are in ω(x) because by Proposition 6.4 f(ω(x)) = ω(x). On the other hand any point z orb x \ orb x is in ω(x) because by the definition we will be able to find a sequence of points of orb x converging to z. So, if x is recurrent then orb x = ω(x). On the other hand, if x is not recurrent then by the definition x orb x \ ω(x) and so orb x ω(x), orb x ω(x) which completes the proof. The notion of a recurrent point is important because in fact for every invariant measure µ the set of recurrent points R(f) has full µ-measure. Let us show this using some of the results established in DYNAMICAL SYSTEMS I (generally, in the situation when we need to refer to measure-theoretic results from DYNAMICAL SYSTEMS I we will 13

do this without a proof). In DYNAMICAL SYSTEMS I, Remark 7.21 it is shown that if A X is a subset such that ν(a) = 1 for every ergodic measure ν then µ(a) = 1 for every measure µ. Hence it is enough to prove that ν(r(f)) = 1 for an ergodic measure ν. To this end let us consider the so-called support of the measure ν denoted by supp(ν) and defined as follows: for every invariant measure µ the set supp(µ) is the set of all points x such that for any neighborhood U of x we have µ(u) > 0. PROPOSITION 7.6. The set supp(µ) is closed, invariant and of full µ-measure. Proof. Let us show that A = supp(µ) is closed. Indeed, if y A then there exists an open set U such that µ(u) = 0. Hence by the definition U is contained in the complement of A which implies that this complement is open and hence A is closed. Let us show that µ(a) = 1. Indeed, every point x A has a neighborhood U such that µ(u) = 0. Since X is a metric compactum it has a countable basis, so we can now represent the set X \A as a countable union of open sets with zero measure which implies that µ(x \ A) = 0 as desired. Let us show that A is invariant. Indeed, let x A while f(x) A. Then for some open set U containing f(x) we have that µ(u) = 0. Therefore, µ(f 1 (U) = 0 as well while f 1 (U) clearly is a neighborhood of x, a contradiction with the assumption that x A. Hence f(x) A and A is invariant. PROPOSITION 7.7. Let ν be an invariant ergodic measure and let B = supp(ν). Then ν-a.e. point x B is such that ω(x) = B (and so all these points are recurrent). Proof. By the construction every point y B is such that for every open set U containing y we have ν(u) > 0. Since ν(b) = 1 by Proposition 7.6 then ν(y ) = ν(y B) for any set Y. So, for any open U such that U B we have ν(u B) > 0. Therefore by DYNAMICAL SYSTEMS I, Corollary 7.25 we conclude that the orbit of almost every point of B is dense in B as desired. THEOREM 7.8. For every invariant measure µ we have µ(r(f)) = 1. Proof. It is enough to prove that if D is a set consisting of non-recurrent points then for every µ we have µ(d) = 0. By Proposition 7.7 this is the case if ν is ergodic. In other words, the ergodic measure of D is always zero. By DYNAMICAL SYSTEMS I, Ergodic Decomposition Theorem (Theorem 7.15) we conclude that then µ(d) = 0 for every measure µ as desired. 14

8 The Center of a Dynamical System and its Depth The closure R(f) of the set R(f) of all recurrent points of a map f is called the center of the map f and is denoted by C(f). There is a different way of defining the set C(f). Namely, it can be obtained by taking Ω(f), then Ω(f Ω(f) ), etc continuing by means of the so-called transfinite induction until the sequence stabilizes. Observe that since on each step the obtained set is closed then the resulting set C(f) is closed as well. Let us verify that the set R(f) is contained in the result of the above described inductive process. If Y X is constructed at some point then on the next step we construct the set Ω(f Y ) (the initial set is Y = X). Indeed, if x Y is a recurrent point then there exists a sequence {n k } such that f n k (x) x. Therefore for any open U containing x we have that f n k (x) U for big enough k which implies that x Ω(f Y ). Since nothing in the above argument depends on Y we see that x C(f) and therefore R(f) C(f). Since C(f) is closed we conclude that indeed R(f) C(f). It was proven by Birkhoff that the resulting set coincides with R(f). The number of steps one has to make in order to get the set C(f) is called the depth of the center C(f) of the map f. As Example 5.2 shows, C(f) can be smaller than Ω(f), so the notion of the depth of the center of a map makes sense. Our main aim now is to study the center of an interval map. A useful tool for us will be the following lemma. LEMMA 8.1. Let U be a complementary to Per(f) interval. Then one of the following cases takes place: (1) points of U never come back into U; (2) for any x U, n such that f n (x) U we have x < f n (x); (3) for any x U, n such that f n (x) U we have x > f n (x). Proof. Let x U, n are such that f n (x) U and x < f n (x). Then since there are no periodic points of f between x and f(x) we conclude that for any m the direction in which f m maps x and the direction in which it maps f n (x) are the same. In other words, either 1) f m (x) > x and f n+m (x) > f n (x), or 2) f m (x) < x and f n+m (x) < f n (x). Let us apply this to m = n; this yields that since f n maps x to the right, f n must map f n (x) to the right as well, and so x < f n (x) < f 2n (x). Hence, f 2n maps x to the right, and applying the above claim to f 2n we see that since f 2n maps x to the right, f 2n must map f n (x) to the right as well and so x < f n (x) < f 3n (x). Clearly, repeating this argument we see that x < f kn (x) for any k 1 and therefore for any z U we have z < f kn (z). 15

Now, suppose that there exists another point y U and a number r such that f r (y) U and f r (y) < y. Similarly to the previous paragraph we see that then f jr (y) < y for any j 1 and therefore for any z U we have f jr (z) < z. Now, let us take s = nr and consider f nr (z) and its position with respect to z. By the previous paragraph, f nr (z) < z but by the paragraph before that f nr (z) > z, a contradiction which proves that the assumptions of the above two paragraphs cannot coexist and completes the proof. An informal way to state Lemma 8.1 is to say that points of U return to U to the same side of themselves; this clearly means that there are right intervals U and there are left intervals U. Let us now prove the following theorem which relies upon Lemma 8.1. THEOREM 8.2. For a continuous interval map f we have C(f) = Per(f). Proof. Let U be a complementary to Per(f) interval. Suppose that x U is recurrent. Then x has to come back to itself closer and closer under growing to infinity sequence of iterations of f. Now, suppose that the least n with f n (x) U is such that x < f n (x). Then, as Lemma 8.1 shows, U is the right interval and f k (z) > z for every z U and k such that f k (z) U. For every m n such that f m (x) U we have by the choice of n that m > n and therefore f m (x) > f n (x) which implies that f m (x) cannot converge to x even along a subsequence. Similar arguments in the case of a left interval U finally imply that x cannot be recurrent as desired. Let us now study the depth of interval maps. To begin with let us give some examples. First, let f : [0, 1 [0, 1] be the identity map. Then since Ω(f) = [0, 1] the depth of the center is 0. Now, consider any orientation preserving homeomorphism f which is not the identity map. Then Ω(f) = Per(f) [0, 1] and it is easy to see that Ω(f Ω(f)) = Ω(f). In other words, the sequence of sets I, Ω(f), Ω(f Ω(f)),... stabilizes after exactly one step, and therefore the depth of f is 1. EXERCISE 8.3. Let f : [ 1, 4] [ 1, 4] be defined as follows: f( 1) = 1, f(0) = 0, f(1) = 3, f(2) = 1, f(3) = 0, f(4) = 1 and linearly on all the intervals into which [ 1, 4] is divided by these points. Show that then Ω(f) Ω(f Ω(f)) by suggesting a point x such that x Ω(f) while on the other hand x Ω(f Ω(f)). By Exercise 8.3 there are interval maps whose depth is greater than 1. However, as we will show next time it cannot be greater than 2. Thus, the map f from Exercise 8.3 has to be of depth exactly 2. Preparing for the next lecture you are welcome to think about how this can be proven directly, yet this is not a part of your homework. 16

9 The Depth of a Dynamical System on the Interval In this lecture we will assume that f : [0, 1] [0, 1] is a continuous interval map. LEMMA 9.1. Let U be an interval containing no periodic points. Then for any x [0, 1] the points of the trajectory {x, f(x),...} which belong to U, form a strictly monotonic (finite or infinite) sequence. Proof. Assume that U is right, {x, f(x),...} U > 1, and m < n are the least with f m (x), f n (x) U. By Lemma 8.1 f m (x) < f n (x), and for any k > n with f k (x) U we get f n (x) < f k (x). Repeating the argument proves the lemma. COROLLARY 9.2. Let U be an interval containing no periodic points. If x Ω(f) U then x never comes back into U. Proof. Assume that U is a right interval. Let x Ω(f) U and contrary to the claim there exists n such that f n (x) U. Then x < f n (x) and we can choose a small neighborhood V of x such that v < f n (V ), V f n (V ) =. Clearly this implies that for all m > n we have f m (V ) V = and x is wandering, a contradiction. THEOREM 9.3. For an interval map f we have Ω(f Ω(f)) = Per(f) = C(f) and so the depth of an interval map cannot be greater than 2. Proof. By Corollary 9.2 if x Ω(f) U where U is complementary to Per(f) then x never comes back into U. Therefore, U Ω(f) is wandering in Ω(f) and hence contains no points of Ω(f Ω(f)). On the other hand, Per(f) Ω(f Ω(f)) as was proved in the previous lecture. Therefore, indeed Ω(f Ω(f)) = Per(f) = C(f) as desired. PROPOSITION 9.4. Let U be an interval containing no periodic points. Then U ω(x) consists of at most one point for any x. Proof. Follows immediately from Lemma 9.1. Now we show that the set Ω(f Ω(f)) U is at most countable. Let us denote the set i=0 f i (A) by orb f (A) = orb(a). The claims in Lemma 9.5 are for students to establish on their own and must be proven in order to understand the lemma. LEMMA 9.5. Let U be an interval containing no periodic points, x Ω(f) U. Then there exists a wandering semi-neighborhood of x. Proof. Without loss of generality assume that U = (a, b) is a right interval and show that there exists an ε > 0 such that I = (x, x + ε) is wandering (i.e., f n (I) I = for any n > 0). By way of contradiction assume that this is false. CLAIM 9.6. Show that under the assumption x orb(f(i)) for every ε > 0. Choose a very small ε > 0 and consider A = orb(f(i)). 17

CLAIM 9.7. Show that there exists n > 0 such that f n (f(i)) f(i). The set A is a countable union of intervals. By Exercise 9.7 there exists n > 0 such that f n (f(i)) is not disjoint from f(i). Denote f(i) by M and consider orb f n(m) = A 0; in this union of intervals two consecutive intervals are non-disjoint, hence A 0 is an interval of some kind. Then A 1 = f(a 0) = f(orb f n(m)) = orb f n(f(m)) is therefore an interval itself, and so are its further images A 2 = f 2 (A 0),..., f n 1 (A 0) = A n 1. It is clear that f n (A 0) = orb f n(f n (M)) A 0, and so intervals A 0,..., A n 1 are cyclically permuted by f. EXERCISE 9.8. Show that connected components of A are cyclically permuted by f. Since U is a right interval, points of I can be mapped into U only to the right of x. Hence, (a, x) A =. By Claim 9.6, x A, hence there exists a unique component B of A with x B. By Lemma 8.1 x A, hence x B. Thus, B is an interval with the left endpoint x / B. Since by Claim 9.8 components of B are cyclically permuted by f, we can find the least N such that B, f(b),..., f N 1 (B) are disjoint while f N (B) B. Let us show that then x f N (B). Indeed, otherwise there exists a small α > 0 such that L = [x, x + α] I B is disjoint from f N (B). Since other components of A are disjoint from B we see that L is disjoint from A. On the other hand, orb(f(l)) orb(f(i)) = A which implies that L is wandering, a contradiction. Thus, there exists y B such that f N (y) = x. Since x is not periodic, y x. In fact, y cannot belong to B. Indeed, if so then y belongs to some iterate of I and hence there are points of I which are eventually mapped into x. However this is impossible since U is a right interval. So, y / B and hence y is the right endpoint of B = (x, y). We showed that y is the unique f N -preimage of x in B. Assume that y f N (B). Then x f 2N (B). Choose α > 0 so that L = [x, x + α ] I B is disjoint from f 2N (B). Then as before L is wandering, a contradiction. Thus, y f N (B). If y = f N (x) then x is periodic, a contradiction. So, y has preimages inside B which again means that some points of I are eventually mapped into x, a contradiction which completes the proof. THEOREM 9.9. The set Ω(f) \ Per(f) is at most countable. Proof. (suggested by J. Malaugh) Consider a complementary to Per(f) right interval U and associate semi-neighborhoods to points x Ω(f) U as in Lemma 9.5. If V and V are such neighborhoods and V V then one of them contains the left endpoint of the other, a contradiction with the fact that they are wandering as sets while their left endpoints belong to Ω(f). Hence, all such neighborhoods are pairwise disjoint which implies that there are no more than countably many of them and completes the proof. 18

10 Transitive Maps 1 Under some assumptions we have Ω(f) = X. E.g., define an n-saw map h : [0, 1] [0, 1] so that points 0, 1/n,..., 1 are mapped alternatively into 0 or 1 starting with h(0) = 0 while in-between these points the map is defined a linear. LEMMA 10.1. For any saw map f we have Ω(f) = [0, 1]. Proof. Assume that f is an m-saw map. Consider any interval I and show that its image is the entire [0, 1]. Indeed, for any interval J the length of its image f(j) = m J if J contains no turning points of f. Since this cannot go on forever there is n such that f n (I) contains a turning point of f and hence f n+1 contains 0 or 1. As follows from the definition, 0 is always fixed, and 1 is either fixed or mapped onto 0. Hence, f n+2 (I) contains a fixed endpoint of [0, 1], and without loss of generality we can assume that 0 = f(0) f n+2 (I). The interval f n+2 is then expanded by f until its image covers 1/m, and then the next image is [0, 1] as desired. This implies that there is a periodic point inside any interval. So periodic points are dense and hence Ω(f) = [0, 1]. A map f : X X is topologically exact if for any open U there is n such that f n (U) = X. CLAIM 10.2. If f : X X is topologically exact then Ω(f) = X. Proof. The claim follows from the fact that for f there are no wandering open sets. We say that f : X X is transitive if there exists x X such that ω(x) = X (in which case we say that x has a dense orbit in X). The existence of a dense in X orbit is an important property. It can be characterized also in a different way. We are going to use in the proof of equivalence of various characterizations the Baire category method. Any compact metric space is complete, that is every Cauchy sequence is convergent. In such a space the Baire Theorem holds: the intersection of a countable family of open dense sets is dense. The intersection of a countable family of open sets is called a G δ -set. The sets containing dense G δ -sets are in some sense large. They are called residual sets (they complements are called the sets of first category). Thus, every open dense set is residual, and by Baire Theorem the intersection of a countable number of residual sets is residual. Paradoxically, often it is much easier to prove that a set is residual than just that it is nonempty. THEOREM 10.3. Let f : X X be a continuous map of a nonempty compact metric space X into itself. Then the following properties are equivalent. 19

(a) f is transitive. (b) The set of points with the dense orbit is residual. (c) For every open sets U, V X there exists n 0 with f n (U) V. (d) For every open sets U, V X and m 0 there exists n m with f n (U) V. Proof. Clearly, (b) implies (a) and (d) implies (c). We will show that (a) implies (d), (c) implies (d), and (d) implies (b). Then it will follow that all four conditions are equivalent. Let us assume (a) and prove (d). Let x be a point with the dense orbit, let U, V X be nonempty open sets, and let m 0. There exists k such that f k (x) U and l k + m such that f l (x) V. Then f n (U) V for n = l k m. Now let us assume (c) and prove (d). We start by proving that f is a surjection (that is, maps X onto itself). Suppose that f(x) X. Take a point x X \ f(x) and a point y f(x). Since x y we can find their neighborhoods U x and V y which are disjoint so that U V =. Since f(x) is compact, the set X \ f(x) is open, and U may be assumed to be contained in X \ f(x). Since f n (V ) f(x) for every n > 0 we conclude that f n (V ) U = for every n > 0, and so in fact f m (V ) U = for any m 0, a contradiction. Let U, V X be nonempty open sets and let m 0. The set W = f m (V ) is open, and since f is a surjection, it is nonempty. By (c), there exists n 0 such that f n (U) W. Then f n+m (U) V. This proves (d). Finally, let us assume (d) and prove (b). The space X has a countable open base (U i ). That is, there exists a sequence of open sets U i such that for any open set V one can find an open set U j V. Indeed, pick a sequence of finite covers of X by 1/n-balls and set U i as the sequence of elements of these covers. If V is open then we can find a ball B(y, ɛ) inside V. Choose n so that 2/n < ɛ and then a ball U i of radius 1/n containing y. Obviously, U i V. Let A i,j be the set of those points x X for which there exists n j such that f n (x) U i. In other words, A i,j = n=j f n (U i ). Since f is continuous, A i,j is open. We will show that it is dense. Indeed, if A i,j is not dense then there exists an open nonempty set V disjoint from it. That is, f n (V ) U i = for every n j. However, this contradicts (d). Therefore A i,j is dense for all i, j. By the Baire Theorem, the set A = i,j A i,j is residual. If x A then the orbit of x passes through each U i infinitely often. This means that this orbit is dense. Hence, (b) is proved. From Theorem 10.3 it follows immediately that for a transitive map the set of nonwandering points is the whole space. 20

11 Transitive Maps 2 The identity on any space consisting of more than one point is a simple example of a map where the set of non-wandering points is the whole space but that is not transitive. Clearly, by Theorem 10.3 all topologically exact maps are transitive. The opposite is not necessarily true as follows from the following example. EXAMPLE 11.1. Consider the map f : [0, 1] [0, 1] defined as follows: (1) two double-sided sequences... l 1 < l 0 < l 1 <... and... r 1 < r 0 < r 1 <... are chosen so that lim n a n, b n = 0 and lim n = 1 and the points of the sequences alternate (i.e. l i < r i < l i+1 for any i); (2) the map f is defined on L = {l i } as follows: f(l i ) = l i 1 ; (3) the map f defined on R = {r i } as follows: f(r i ) = r i+1 ; (4) the map f at 0, 1 is defined as follows: f(0) = 0, f(1) = 1; (5) the map f is defined linearly on every interval complementary to R L 0, 1. Show that f is transitive but not topologically exact. Proof. It is easy to see that by the definition 0, 1 do not have preimages inside (0, 1). Hence the map f cannot be topologically exact. To show that it is transitive by Theorem 10.3 we need to prove that for any interval I its orbit is dense in [0, 1]. Observe that by the definition f is expanding on every interval of monotonicity (has the slope with absolute value greater than 1). Now, let us first prove that there are two images of I, say, f n (I) and f m (I) which are not disjoint. To this end we study the length of the iterates of I depending on whether they intersect L R or not. Consider two cases. First, assume that iterates of I never intersect L R. Then since L R are exactly the turning points of f and iterates of I do not cover any of them we see that the length of images of I grows. Thus, we get a sequence of intervals with lengths bounded from below by I. Clearly, some of them must intersect, and the claim is proven. Now, suppose that for some k we have f k (I) (L R). Without loss of generality we can assume that r 0 f k (I) and there exists a non-degenerate interval J I with the right endpoint r 0. If we can choose J so that it covers l 0 then its image will intersect itself and hence I will have two non-disjoint images. If J cannot be chosen so that it contains l 0, let us consider its forward images and show that there exists i such that l i, r i f i (I). Indeed, otherwise I < f(i) <..., so if i is chosen in such a way that r i l i < I we see that I i (whose right endpoint is r i ) must stretch from r i all the way to the left to 21